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Revision Sheet for Grade 12 Final Term II 85% of your Final exam mark is from this sheet.

1. Arrange the following molecules in increasing order of boiling point: HF, HCl, HBr, HI. Explain your answer.

HCl, HBr, HI, HF HF is the only one exhibiting H bonding. All the others are bonded by Van der Waal forces (dipole-dipole). H-bonds are stronger, so HF will have the highest boiling point. HCl is the smallest, HI is the largest, Van der Waal forces depend on the number of electrons present in the molecules. Since HI has the largest number of electrons it will have the strongest Van der Waal forces. And hence the second highest boiling point after HF and so on.

2. Identify the major intermolecular forces that exist in the following molecules: HCl, NH3, CO2, CH4.

HCl : dipole-dipole NH3 : H – bonding CO2 : London dispersion CH4 : London dispersion

3. Account for each of the following: a) At 25oC and 1 atm, F2 is gas, whereas I2 is solid.

Both have the same type of intermolecular forces, namely Van der Waal forces. Van der Waal forces increase in strength with the increase in the number of electrons and protons present. I2 is larger and so has more electrons and protons. I2 has much stronger VDW making it a solid whereas F2 is a gas.

b) The melting point of MgO is greater than the melting point of CsCl Lattice energy of MgO is greater than CsCl since both the anion and cation in MgO carry twice the charges on anion and cation of CsCl. Hence, more energy is required to break the MgO crystal than CsCl crystal, so MgO has a greater melting point.

c) Ammonia, NH3, is very soluble in water, whereas phosphine, PH3 is only moderately soluble in water.

Ammonia is able to H-bond with water so it is highly soluble. Phosphine cannot H-bond, but is polar. Therefore, it will be slightly soluble. d) Si melts at a much higher temperature than I2.

Si forms a network solid like carbon. The large number of covalent bonds that must be broken to liquefy the solid requires a very high temperature. In I2, there is VDW

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forces which are relatively much weaker and hence can be broken at a much lower temperature.

4. a. Draw a phase diagram for water. Identify on the diagram the following:

The solid, liquid and gas regions; the solid–liquid, solid–gas, and liquid–gas equilibrium lines; the triple and critical points. Solid - liquid line Critical point Liquid Liquid - gas line Pressure Solid Gas Triple point Solid - gas line temperature b. How does the melting point of ice vary with the pressure?

As the pressure increases the melting point decreases, since the solid-liquid equilibrium line has a negative slope.

5. The compound pentane, C5H12, occurs in three isomeric forms, n-pentane, 2-

methylbutane and 2,2-dimethylpropane. (a) Draw the structural formula of each of the isomers (include all atoms). Clearly label each

structure. H H H H H H - C – C – C – C – C –H n-pentane H H H H H H H H H H - C – C – C – C – H 2-methylbutane H H H H – C – H H

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H H – C – H H H H – C – C – C – H 2,2-dimethylpropane H H H – C – H H 10.1 Identify molecule with highest boiling point in a list

(b) On the basis of molecular structure, identify the isomer that has the highest boiling point. Justify your answer. The isomer n-pentane has the highest boiling point. In all isomers, the intermolecular forces are London (dispersion) forces These are greatest in n-pentane because molecules of n-pentane, with its linear structure, can approach one another more closely and can form a greater number of induced temporary dipoles than the other isomers which have more compact structure.

6. 10.0g sugar (molar mass = 342) were dissolved in 100g H2O. Calculate the mass % of the

sugar in the solution. Given: msugar = 10.0g, mwater = 100g RTF: mass %

mass % of sugar = = 10 ×100110

= 9.1%

7. 10.0g sugar (molar mass = 342) were dissolved in 100g H2O. Calculate the mole fraction

of the sugar in the solution. Given: msugar = 10.0g, mwater = 100g RTF: Xsugar

Xsugar = = -3

10342 = 5.24×10100 10+

18 342

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8. 10.0g sugar (molar mass = 342) were dissolved in 100g H2O to give a final volume of

102cm3. Calculate the molality of the sugar in the solution. Given: msugar=10.0g, mwater = 100g RTF: molality

Nsugar = 10 = 0.0292mole342

m = = -3

0.0292100×10

= 0.292 molal

9. 10.0g sugar (molar mass = 342) were dissolved in 100g H2O to give a final volume of

102cm3. Calculate the molarity of the sugar in the solution. Given: msugar=10.0g, V = 102 cm3 RTF: molarity

nsugar = 10 = 0.0292mole342

[sugar] = 0.0292 = 0.287M0.102

10. What is the molality of ethanol, C2H5OH, in an aqueous solution in which the mol

fraction of ethanol is 0.082? Given: Xethanol = 0.082 RTF: molality

X =

0.082 = m = 4.9767 = 5.0 molal

11. An HCl solution is 30% HCl by weight and its density is 1.1g mL-1. What is its molarity?

Given: mass % HCl = 30, dsolution = 1.1 g/ml RTF: Molarity Assume Vsolution = 1000ml msolution = (dsolution)(Vsolution) = (1.1)(1000) = 1100g

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mHCl = (1100) = 330g nHCl = = 9.04 moles

[HCl] = = 9.04M

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12. 5.85g NaCl were dissolved in 32.4g of water at a temperature where the vapour pressure

of the water is 25mmHg. Calculate the vapour pressure of the solution. Given: mNaCl = 5.85g, mwater = 32.4g, VPwater = 25mm Hg RTF: P solution

Psolution = Xwater = + -

° °water waterwater water

total waterNa Cl

n nP = Pn n + n + n

⎛ ⎞⎜ ⎟⎝ ⎠

soln

32.418P = × 25 = 22.5 mmHg

5.85 32.42 +58.5 18

13. Acetone and chloroform form solutions that exhibit negative deviations.

a. Represent the behavior of these two liquids on a diagram of vapour pressure vs. mole fraction of each component of the solution. Vapour Pressure 0 --------- XA ------- 1

b. What can you say about the acetone – chloroform attractions as compared to the acetone – acetone or chloroform – chloroform attractions?

stronger

c. Compare the actual total pressure of the solution with the expected (calculated) pressure. P calc > P actual

14. 10.0g sugar (molar mass = 342) were dissolved in 250g of water. Calculate the boiling point of the solution. Kb = 0.51.

Given: msugar = 10 g, msolvent = 250 g, Kb = 0.51 RTF: boiling point

molality =

1 03 4 2 0 .1 1 7

0 .2 5 0m=

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∆Tb = mKb = (0.117)(0.51) = 0.0596°C ⇒ B.P. = 100.0597◦C

15. 0.50g of a compound when dissolved in 40g of CHCl3 raised the boiling point by 0.40°C. What is the molecular weight of the compound? Kb = 3.6.

Given: mcmpd = 0.50g, msolvent = 40g, ΔTb = 0.40°C, Kb = 3.6 RTF: Mcompound

∆Tb = mKb m = = 0.11

m = = = 0.11 M = 113.6 g/mol = 1.1 x 102 g/mole

16. 0.5g of ethylene glycol were dissolved in 1dm3 solution at 27 °C. The osmotic pressure of the solution was 147 mmHg. Calculate the molar mass of ethylene glycol.

Given: msolute = 0.5 g, Vsolution = 1 dm3, t = 27 + 273 = 300K, Psolution= 147 mmHg RTF:M π = CRT

C = n m π= =V MV RT

⇒ M = 1⎛ ⎞×⎜ ⎟

⎝ ⎠

mRT (0.5)(0.082)(300)= =147πv760

63.6 g/mol

17. 0.585g NaCl were dissolved in 100g of water. Calculate the freezing point of the

solution. Assume that NaCl dissociates completely. Kf = 1.86. Given: mNaCl = 0.585 g, mwater = 100 g, Kf = 1.86, i = 2 RTF: freezing point

∆Tf = miKf = iKf = iKf = (2)(1.86) = 0.327°C Freezing point = - 0.372 ◦C

18. Consider the following reaction: N2 + 3H2 → 2NH3 The rate of the reaction was studied by introducing 6 moles of H2 and 4 moles of N2. After 30mins, the following number of moles was found to exist: 5.1 moles of H2, 3.7 moles of N2 an 0.6 moles of NH3. Calculate the rate of the reaction in terms of the variation of concentration of each chemical with time.

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N2 + 3H2 → 2NH3

Initially 4mol 6mol After 30 min 3.7mol 5.1mol 0.6mol

Rate with respect to N2 = 4 3.730−

= 0.010 mole / min

Rate with respect to H2 = 6 5.130−

= 0.030 mole / min

Rate with respect to NH3 = 0.630

= 0.020 mole / min

19. For the reaction: 2X + Y → Z, the following results were obtained:

__________________________________________________________________ Experiment [ Xo ] [ Yo ] Initial Rate __________________________________________________________________ 1 0.3 0.8 2.4 2 0.3 1.6 9.6 3 0.1 1.6 3.2 __________________________________________________________________ a. What is the rate law for this reaction?

Consider experiments 1+2: [X]=same

As [Y] doubles, rate quadruples ⇒rateα [Y] 2

Consider experiments 2+3: [Y]=same As [X] decreases by 3 times, rate decreases by 3 times ⇒rateα [X] Rate = k[Y] [X] 2

b. Calculate the rate constant of the reaction.

Given: Rate = k[Y] [X] 2

RTF: k Rate = k[Y] [X] choose any experiment, let us say experiment 1

2

2.4 = k (0.8)2(0.3) k = 12.5

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20.

a. The rate of the reaction: 2X → X2, was studied. The following results on the concentration of X were obtained:

__________________________________________________________________ [ X ] 1 0.8 0.6 0.4 0.2 Time (s) 0 10 23 41 72 __________________________________________________________________ Show that the reaction is first order with respect to X. What are the value and units of the specific rate constant? Must plot ln[X] versus time. You get a straight line with a negative slope which shows that the reaction is first order. OR Plot [X] versus time. Calculate two half-lives. They will be the same which shows that the reaction is first order.

b For the reaction: 2X + Y → 2Z, the initial rate was studied and the following results were obtained:

__________________________________________________________________ Experiment [ Xo ] [ Yo ] Initial Rate __________________________________________________________________ 1 0.2 0.2 2.4 2 0.2 0.3 2.4 3 0.4 0.1 9.6 __________________________________________________________________ What is the rate law for this reaction? Consider experiments 1 and 2: [X] = same As [Y] changes the rate remains the same ⇒ [Y] does not affect the rate Rate = k[X]n Consider experiments 1 and 3: Rate3 = k Rate1 = k Divide the two equations:

4 = 2n 22 = 2 n ⇒ n = 2 Rate = k[X]2

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21. The decomposition of a certain chemical, X2O, is first order with a constant of 1.4 x 10-2s–

1. a. Determine the half–life for the decomposition of X2O. Given: k=1.4 × 10-2 s-1 , first order reaction RTF: half-life

1 -22

ln2 0.693t = = = 49.5seck 1.4×10

b. The initial pressure of a sample of X2O was 6atm. How long will it take the sample to decompose and reach a pressure of 0.75atm? Given: Pinitial = 6atm, Pfinal = 0.75atm RTF: time

⇒

n0

n 3

1m = m ( )2

0.75 1 1= ( ) = 0.125 = ( ) n = 36 2 2

t = 3(49.5) = 148.5sec

OR lnPf = -kt + lnP° ln0.75 = (-1.4 × 10-2)t + ln 6 t = 148.5 sec

22. CH4(g) + 2Br2(g) → CH2Br2(g) + 2HBr(g) Methane gas reacts with bromine gas to form dibromomethane and hydrogen bromide, as represented by the equation above.

(a) A 50.0 g sample of methane gas is placed in a reaction vessel containing 5.16 mol of Br2(g).

(i) Identify the limiting reactant when the methane and bromine gases are combined.

Justify your answer with a calculation. Give: mmethane = 50.0g, nbromine = 5.16 mole RTF: limiting reactant CH4(g) + 2Br2(g) → CH2Br2(g) + 2HBr(g) 16g 2moles 50.0g ? ? = 6.25 moles

Br2 is the limiting reactant because, in order to react with the given amount of CH4, more moles of Br2 are required than the 5.16 moles of Br2 that are present.

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(ii) Calculate the total number of moles of CH2Br2(g) in the container after the

limiting reactant has been totally consumed. Give: mmethane = 50.0g, nbromine = 5.16 mole RTF: n of CH2Br2 CH4(g) + 2Br2(g) → CH2Br2(g) + 2HBr(g) 2moles 1mole 5.16 moles ? ? = 2.58 moles Initiating most reactions involving bromine gas involves breaking the Br–Br bond, which has a bond energy of 121 kJ mol−1. (b) Calculate the amount of energy, in joules, needed to break a single Br–Br bond. Given: Br2 bond energy = 121 kJ/mol RTF: Br2 bond energy per bond

121 kJ × = 2.01 × 10−19 J.

(c) Calculate the longest wavelength of light, in meters, that can supply the energy per photon necessary to break the Br–Br bond. Given: E = 2.01 × 10−19 J RTF: λ

E = h = 9.89 × 10−7 m

The following mechanism has been proposed for the reaction of methane gas with bromine gas. All species are in the gas phase. Step 1 Br2 2Br• fast equilibrium Step 2 CH4 + Br• → •CH3 + HBr slow Step 3 •CH3 + Br2 → CH3Br + Br• fast Step 4 CH3Br+ Br• → CH2Br2 + H• fast Step 5 H• + Br• → HBr fast (d) In the mechanism, is CH3Br a catalyst, or is it an intermediate? Justify your answer. CH3Br is an intermediate because it is produced in step 3 and consumed in step 4 of the reaction mechanism.

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(e) Identify the order of the reaction with respect to each of the following according to the

mechanism. In each case, justify your answer. (i) CH4(g) All steps in a reaction mechanism are elementary steps. The rate laws of elementary steps can be predicted from the reaction stoichiometry. The rate law of the rate determining step is the rate law of the overall reaction. Rate = k[CH4][Br•] But the first step is a quick equilibrium relating the [ ] as follows:

Keq = [Br•] = (Keq[Br2])1/2 ⇒ Rate = k(Keq[Br2])1/2 [CH4] The order of the reaction with respect to CH4 is 1.

(ii) Br2(g) The order of the reaction with respect to Br2 is 1/2

23. H2(g) + Br2(g) → 2HBr(g) The table below gives data for a reaction rate study of the reaction represented above.

Experiment Initial [H2]/ (mol L−1)

Initial [Br2] / (mol L−1)

Initial Rate of Formation of HBr /

(mol L−1 s−1)

1 0.00100 0.000500 3.64 × 10−12 2 0.00200 0.000500 7.28 × 10−12 3 0.00200 0.000250 3.64 × 10−12

(a) Determine the order of the reaction with respect to H2 and justify your answer.

The order of the reaction with respect to H2 is 1. Comparing experiments 1 and 2, doubling the initial concentration of H2 while keeping the initial concentration of Br2 constant results in a doubling of the reaction rate. (b) Determine the order of the reaction with respect to Br2 and justify your answer.

The order of the reaction with respect to Br2 is 1. Comparing experiments 2 and 3, halving the initial concentration of Br2 while keeping the initial concentration of H2 constant results in a halving of the reaction rate. (c) Write the overall rate law for the reaction.

rate = k [H2][Br2]

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(d) Write the units of the rate constant. M-1s-1

(e) Predict the initial rate of the reaction if the initial concentration of H2 is 0.00300 mol L−1

and the initial concentration of Br2 is 0.000500 mol L−1. Given: [H2] = 0.00300 mol L-1, [Br2] = 0.000500 mol L-1 RTF: rate For this reaction, the initial concentration of Br2 is the same as in Experiment 1 but the initial concentration of H2 is three times as large. And because the reaction is first order with respect to each reactant, the initial rate of the reaction would be 10.92 × 10−12 mol L−1 s−1, which is three times the rate of the initial rate of the reaction in Experiment 1. OR Use experiment 1 to find k Rate = k[H2][Br2]

k = = 7.28 × 10-6 Rate = (7.28 × 10-6)(0.00300)(0.000500) = 10.92 × 10−12 mol L−1 s−1

24. Sr-90 has t1/2 = 29 years. How long does it take 80% of a sample of Sr to decay?

67 years

25. Kp for NH4Cl(s) NH3(g) + HCl(g) is 1.00 x 10-2 atm2 at a particular temperature.

What are the partial pressures of NH3 and HCl?

NH4Cl(s) NH3(g) + HCl(g) Pinitial ---- --- --- Change in P -p +p +p Peqbm ---- p p

Kp = PHCl 1.00 × 10-2 = p2 p = 0.100 atm = PHCl = 0.100 atm

26. I. This question concerns the reaction: SO2(g) + ½ O2(g) SO3(g) ΔH < 0

Some SO2, O2, and SO3 gases exist at equilibrium in a closed container. Predict the effect of the following changes on the quantity of SO3. a. Increasing the temperature. Given: Exothermic reaction, increase in temperature

RTF: Effect on SO3 quantity An increase in temperature favours the endothermic reaction. The reaction is exothermic so the equilibrium position will shift to the left. Hence, the quantity of SO3 will decrease.

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b Reducing the volume of the container. Given: decrease in volume

RTF: Effect on SO3 quantity A decrease in the volume increases the pressure. The equilibrium position will shift to the right where the number of moles of gases is less. Hence, the quantity of SO3 will increase. c Some catalyst is added. Given: adding a catalyst

RTF: Effect on SO3 quantity A catalyst does not affect the equilibrium position. So, the quantity of SO3 will not change.

What is the effect of adding NaOH(aq) to an equilibrium system containing saturated Cu(OH)2 solution? Given: Adding NaOH RTF: Effect on [ ] Cu(OH)2(s) Cu2+(aq) + 2OH-(aq) Adding NaOH(aq), increases [OH-]. To decrease the [OH-], the system shifts to the left, the side that consumes OH-. Hence, [Cu2+] decrease.

II. What is the effect on the % yield of raising the pressure on the following equilibrium

system: N2(g) + 3H2(g) 2NH3(g) ΔH < 0

Given: increasing pressure RTF: Effect on % yield A pressure increase always favors the side with less moles of gases to decrease it. In this case, equilibrium will shift to the right where there are 2 moles of gases compared to 4 on the reactants side. Since more ammonia is produced, then % yield increases.

III. What is the effect on the % yield of raising the temperature on the following equilibrium

system: N2(g) + 3H2(g) 2NH3(g) ΔH < 0

Given: increasing temperature, exothermic reaction RTF: Effect on % yield An increase in temperature favors the endothermic reaction. The reaction is exothermic so the equilibrium position will shift to the left. Hence, the quantity of NH3 will decrease which decreases the % yield.

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IV. Consider the reaction: SO2(g) + O2(g) SO3(g) ΔH < 0 At particular temperature the equilibrium constant, K, has a value of 10. Explain using Le Chatelier’s Principle, what will happen to the value of K if the temperature is increased. Given: increasing tempreature RTF: Effect on value of K An increase in temperature favours the endothermic reaction. The reaction is exothermic so the equilibrium position will shift to the left. Hence, the value of the equilibrium constant will decrease.

27. Calculate the [H+] and [OH–] of water at a temperature where the Kw = 4.0 x 10 – 14. Given: Kw = 4.0 x 10 –14 RTF: [H+] and [OH–] Kw = [H+][OH–] = 4.0 x 10 –14 Every time water dissociates it produces equal amounts of H+ and OH- ⇒ [H+] = [OH–] = = 2.0 x 10 –7M

28. Calculate the [H+] and [OH–] in the following solutions.

a) 0.02M HCl(aq) Given: [HCl] = 0.02M RTF: [H+] and [OH–] HCl is a strong acid, it dissociates completely ⇒ [HCl] = [H+] = 0.02M

Kw = [H+][OH–] ⇒ [OH-] = = 5 × 10-13M

b) 0.5M NaOH(aq). Given: [NaOH] = 0.5M RTF: [H+] and [OH–] NaOH is a strong base, it dissociates completely ⇒ [NaOH] = [OH-] = 0.5M

Kw = [H+][OH–] ⇒ [H+] = = 2 × 10-14M

29. a. The pH in a lemon juice was found to be 3. Find the [H+] and [OH–] in the solution. Given: pH = 3 RTF: [H+] and [OH–] pH = -log[H+] [H+] = 1.0 × 10-3M

Kw = [H+][OH–] ⇒ [OH-] = = 1.0 × 10-11M b. Find the [H+] and [OH–] in a solution, the pOH of which is 10.

Given: pOH = 10

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RTF: [H+] and [OH–] pOH = -log[OH-] [OH-] = 1.00 × 10-10M

Kw = [H+][OH–] ⇒ [H+] = = 1.00 × 10-4M

30. Find the [H+], [OH–], pH and pOH of a 0.2M HCl(aq) solution. Given: [HCl] = 0.2M RTF: [H+], [OH–], pH, pOH HCl is a strong acid, it dissociates completely ⇒ [HCl] = [H+] = 0.2M

Kw = [H+][OH–] ⇒ [OH-] = = 5 × 10-14M pH = -log 5 × 10-14 = 13.3 pOH = 14.0 – pH = 14 – 13.3 = 0.7

31. HNO2 is a weak acid, with a dissociation constant, Ka, of 5.00 x 10-5. Calculate the concentration of all species present in a 2.00 M HNO2 solution. Given: [HNO2] = 2.00M, Ka = 5.00 x 10-5 RTF: [HNO2], [NO2

-], [H+] ar equilibrium

HNO2(aq) H+(aq) + NO2-

(aq) [initial] 2.00 - -

[change] -x +x +x [eqbm] 2.00 – x x x

Ka = = 5.00 x 10-5 assume x << 2.0

x2 = 1.00 × 10-4 [H+] = [NO2-] = 1.00 × 10-2M (assumption valid)

[HNO2] = 2.00 - 1.00 × 10-2 = 1.99M

32. Find the pH of a 2.0 M HNO2 solution. Given :[H+] = 1.0 × 10-2M (from previous BQ) RTF: pH

pH = -log 1.0 × 10-2M = 2.00

33. HCN is a very weak acid, that is only 5 x 10-3% dissociated in 1 M solution. Calculate the value of the dissociation constant of HCN.

Given: % dissociation = 5 x 10-3, [HCN] = 1M RTF: Ka % dissociation = α × 100 α = 5 x 10-5

Ka = = 2.5 × 10-9

34. Find the [H+], [OH–], pH and pOH of a 0.2M NaOH(aq) solution.

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Given: [NaOH] = 0.2M RTF: [H+], [OH–], pH, pOH NaOH is a strong acid, it dissociates completely ⇒ [NaOH] = [OH-] = 0.2M

Kw = [H+][OH–] ⇒ [H+] = = 5 × 10-14M pOH = -log 5 × 10-14 = 13.3 pH = 14 – pH = 14 – 13.3 = 0.70

35. Classify the following as acidic, basic or neutral salts. CuCl2 This salt is the result of the reaction between the weak base, Cu(OH)2, a n d t h e s t r o n g a c i d HCl. Therefore, it is an acidic salt. K2CO3 This salt is the result of the reaction between the strong base, KOH, a n d t h e w e a k a c i d H 2 C O 3 . Therefore, it is a basic salt. FeSO4 This salt is the result of the reaction between the weak base, Fe(OH)2, a n d t h e s t r o n g a c i d H2SO4. Therefore, it is an acidic salt. Mg(NO3)2 This salt is the result of the reaction between the weak base, Mg(OH)2, a n d t h e s t r o n g a c i d HNO3. Therefore, it is an acidic salt. (CH3COO)2Ca This salt is the result of the reaction between the strong base, Ca(OH)2, a n d t h e w e a k a c i d C H 3 C O O H . Therefore, it is a basic salt. NH4NO3 This salt is the result of the reaction between the weak base, NH3, a n d t h e s t r o n g a c i d HNO3. Therefore, it is an acidic salt.

36. Calculate the pH of 0.20M NaCN. Ka HCN= 6 x 10-10. Given: [NaCN] = 0.20M, Ka HCN = 6 x 10-10

RTF: pH CN-(aq) + H2O(l) HCN(aq) + OH-(aq) [initial] 0.20 - - [change] -x +x +x [eqbm] 0.20 – x x x

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Kb = = = 1.7 x 10-5 assume x << 0.2

x = = 1.84 × 10-3M = [OH-] (assumption valid) pOH = 2.74 pH = 14 – 2.74 = 11.26

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37. Calculate the pH of 0.10M NH4Cl. Kb of NH3 = 1.8 x 10-5.

Given: [NH4Cl] = 0.10M, Kb NH3 = 1.8 x 10-5

RTF: pH NH4

+(aq) + H2O(l) NH3(aq) + H+(aq) [initial] 0.10 - - [change] -x +x +x [eqbm] 0.10 – x x x

Ka = = = 5.6 x 10-10 assume x << 0.1

x = = 7.5 × 10-6M = [H+] pH = 5.12

38. Calculate the pH of 0.10M acetic acid solution in 0.80M HCl(aq) solution. Ka = 1.8 × 10-5.

Given: [CH3COOH] =0.10M, [HCl] = 0.80M, Ka = 1.8 × 10-5 RTF: pH

CH3COOH(aq) CH3COO-(aq) + H+(aq) [initial] 0.10 - 0.80 [change] -x +x +x [eqbm] 0.10 – x x 0.80 + x

Ka = = 1.8 × 10-5 assume x << 0.1 x = = 2.25 × 10-6 M [H+] = 0.80 + 2.25 × 10-6 = 0.80M pH = 0.097

39. Calculate the pH of a buffer solution which is 0.2M NaF and 0.3M HF. Given: [NaF] = 0.2M, [HF] = 0.3M , Ka = 7.2 × 10-4 RTF: pH

It is a buffer solution, therefore pH = pKa + log

pH = -log7.2 × 10-4 + log = 3.14 +(– 0.18) = 2.96

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40. 50 cm3 of 0.50 M HNO3 solution were titrated with 0.50 M NaOH solution. Find the

pH of the solution when 60cm3 of base are added. Given: = 50 ml, [HNO3] = 0.50M, [NaOH] = 0.50M, VNaOH = 60ml RTF: pH = (50)(0.50) = 25mmoles nNaOH = (60)(0.50) = 30mmoles n of base in excess = 30 – 25 = 5 mmoles NaOH is a strong base, it dissociates completely. nOH- = nNaOH = 5mmoles

[OH-]excess = = 0.045 M pOH = -log0.045= 1.34 pH = 14 – pOH = 14 – 1.34 = 12.7

41. 50 cm3 of 0.50 M HNO3 solution were titrated with 0.50M NaOH solution. Find the

pH of the solution when 10cm3 of base are added. Given: = 50 ml, [HNO3] = 0.50M, [NaOH] = 0.50M, VNaOH = 10ml RTF: pH = (50)(0.50) = 25mmoles nNaOH = (10)(0.50) = 5mmoles n of acid in excess = 25 – 5 = 20 mmoles

[H+] = [HNO3]excess = = 0.33 M pH = -log0.33 = 0.48

42. 50 cm3 of 0.10 M CH3COOH solution were titrated with 0.10 M KOH solution. Find the pH of the solution when no base has been added yet. Ka = 1.80 × 10-5

Given: = 50 ml, [CH3COOH] = 0.10M, [KOH] = 0.10M, VKOH = 0ml, Ka = 1.80 × 10-5 RTF: pH CH3COOH is a weak acid, it dissociates partially:

CH3COOH(aq) CH3COO-(aq) + H+(aq) [initially] 0.10M - - [cange] -x +x +x [eqbm] 0.10 – x x x

Ka = = 1.80 × 10-5 assume x << 0.10

22 

[H+] = x = = 1.34 × 10-3 M (assumption valid) pH = -log1.34 × 10-3 = 2.87

43. 50 cm3 of 0.10 M CH3COOH solution were titrated with 0.10 M KOH solution. Find the pH of the solution when 10 ml of base has been added. Ka = 1.80 × 10-5

Given: = 50 ml, [CH3COOH] = 0.10M, [KOH] = 0.10M, VKOH = 10ml, Ka = 1.80 × 10-5 RTF: pH = (50)(0.10) = 5mmoles nKOH = (10)(0.10) = 1mmole

CH3COOH(aq) + OH-(aq) → CH3COO-(aq) + H2O(l) Initially 5mmoles 1mmole - After reaction

4mmoles - 1mmole

After the reaction we end up with a solution containing a weak acid and its conjugate base ⇒ buffer solution.

pH = pKa + log = 4.74 – 0.60 = 4.14

44. 50 cm3 of 0.1 M CH3COOH solution were titrated with 0.1 M KOH solution. Find the pH of the solution when 25cm3 base are added. (What is the significance of this stage of titration?)

Given: = 50 ml, [CH3COOH] = 0.10M, [KOH] = 0.10M, VKOH = 25ml, Ka = 1.80 × 10-5 RTF: pH = (50)(0.10) = 5mmoles nKOH = (25)(0.10) = 2.5mmoles

CH3COOH(aq) + OH-(aq) → CH3COO-(aq) + H2O(l) Initially 5mmoles 2.5mmol

e -

After reaction

2.5mmoles - 2.5mmoles

After the reaction we end up with a solution containing a weak acid and its conjugate base ⇒ buffer solution.

23 

pH = pKa + log = 4.74 This point is called the half-way equivalence point. The buffer is at its best capacity and the pH = pKa

45. 50 cm3 of 0.1 M CH3COOH solution were titrated with 0.1 M KOH solution. Find

the pH of the solution when 50cm3 base are added. (What is the significance of this stage of titration?)

Given: = 50 ml, [CH3COOH] = 0.10M, [KOH] = 0.10M, VKOH = 50ml, Ka = 1.80 × 10-5 RTF: pH = (50)(0.10) = 5 mmoles nKOH = (50)(0.10) = 5 mmoles

CH3COOH(aq) + OH-(aq) → CH3COO-

(aq) + H2O(l)

Initially 5mmoles 5mmole - After reaction

- - 5mmoles

[CH3COO-] = = 0.050M After the reaction we end up with a solution containing a weak base CH3COO-

(aq).

CH3COO-

(aq) + H2O(l) CH3COOH(aq) OH-(aq)

[initially] 0.050M - - [change] -x +x +x [eqbm 0.050 – x x x

Kb = = = 5.56 × 10-10 assume x << 0.050

[OH-] = x = = 5.27 × 10-6M (assumption valid) pOH = 5.28 pH = 14 – pOH = 14 – 5.28 = 8.72

46. 50 cm3 of 0.1 M CH3COOH solution were titrated with 0.1 M KOH solution. Find the pH of the solution when 60cm3 base are added.

Given: = 50 ml, [CH3COOH] = 0.10M, [KOH] = 0.10M, VKOH = 60ml, Ka = 1.80 × 10-5 RTF: pH

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= (50)(0.10) = 5mmoles nKOH = (60)(0.10) = 6mmoles n of base in excess = 6 – 5 = 1 mmole KOH is a strong base, it dissociates completely. nOH- = nKOH = 1mmole

[OH-]excess = = 9.09 × 10-3 M pOH = -log 9.09 × 10-3 = 2.05 pH = 14 – pOH = 14 – 2.04 = 11.95

47. HCN(aq) +H2O(l) H3O+(aq) + CN-(aq) Ka=6.2 × 10-10

Hydrocyanic acid HCN(aq) dissociates in water as represented by the equation above a) Write the equilibrium-constant expression for the dissociation of HCN(aq) in water

Ka = b) Calculate the molar concentration of H3O+ in a 0.40M HCN(aq) solution

Given: [HCN] = 0.40M, Ka HCN = 6.2 x 10-10

RTF: [H+]

HCN(aq) + H2O(l) CN-(aq) + H3O+(aq) [initial] 0.40 - - [change] -x +x +x [eqbm] 0.40 – x x x

Ka = = = 6.2 x 10-10 assume x << 0.40

x = = 1.57 × 10-5M = [H3O+] (assumption valid) pH = 4.80

c) 1.5 ml 1.0M NaOH is added to 25 ml of 0.40M HCN(aq). Calculate the pH of the solution.

Given: VHCN = 25ml, [HCN] = 0.40M, VNaOH = 1.5ml, [NaOH] = 1.0M, Ka HCN = 6.2 x 10-10 RTF: pH Calculate the number of moles: nHCN = (25)(0.40) = 10mmoles nNaOH = (1.5)(1.0) = 1.5mmoles

HCN(aq) + OH-(aq) → F-(aq) + H2O(aq) Initially 10 1.5 - Finally 8.5 - 1.5

25 

pH = pKa + log = - log 6.2 x 10-10 + log = 9.21 – 0.75 = 8.46

26 

48. CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

Acetic acid dissociates in water as shown in the equation above. A 25.0ml sample of aqueous solution of 0.100M acetic acid is titrated using standardized 0.100M NaOH.

a) After addition of 12.5 ml of 0.100M NaOH, the pH of the resulting solution is 4.74.

Calculate each of the following. (i) [H+] in solution

Given: = 25 ml, [CH3COOH] = 0.100M, [NaOH] = 0.100M, VNaOH = 12.5ml, Ka = 1.80 × 10-5, pH = 4.74 RTF: [H+] This point is called the half-way equivalence point. At this point the pH = pKa.

[H+] = Ka = 1.80 × 10-5M (ii) [OH-] in the solution Given: [H+] = 1.80 × 10-5M RTF: [OH-]

[OH-] = = 5.56 × 10-10M (iii) The number of moles of NaOH added

Given: [NOH] = 0.100M, VNaOH = 12.5ml RTF: nNaOH nNaOH = (0.100)(12.5 × 10-3) = 1.25 × 10-3 mole (iv) The number of mole of CH3COO- in the solution.

Given: = 25 ml, [CH3COOH] = 0.100M, [NaOH] = 0.100M, VNaOH = 12.5ml, Ka = 1.80 × 10-5

RTF: [CH3COO-] = (25)(0.10) = 2.5mmoles nNaOH = (12.5)(0.10) = 12.5mmoles

CH3COOH(aq) + OH-(aq) → CH3COO-(aq) + H2O(l) Initially 2.5mmoles 1.25mmole - After reaction

1.25mmoles - 1.25mmoles

27 

v) The number of moles of CH3COOH present the solution Given: = 25 ml, [CH3COOH] = 0.100M, [NOH] = 0.100M,

VNaOH = 12.5ml, Ka = 1.80 × 10-5 RTF: [CH3COO-]

= (50)(0.10) = 5mmoles nKOH = (25)(0.10) = 2.5mmoles

CH3COOH(aq) + OH-(aq) → CH3COO-(aq) + H2O(l) Initially 2.5mmoles 1.25mmole - After reaction

1.25mmoles - 1.25mmoles

b) State whether the solution at the equilibrium of the titration is acidic, basic, or neutral.

Explain your reasoning. At the equivalence point, we have a solution of CH3COO-. This ion dissociates in

water as follows: CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq) Since OH- ion are formed then the solution is basic.

In a different experiment, 0.118 g sample of a mixture of solid CH3COOH and solid NaCl is dissolved in water and titrated with 0.100M NaOH. The equivalence point was reached when 10.00 ml of the base solution is added.

c) Calculate the following: (i) The mass in grams of acetic acid solid in the mixture Given: [NaOH] = 0.100M, VNaOH = 10.00ml RTF: m of CH3COOH At the equivalence point, nNaOH = nacid since the base is monobasic and the

acid is monoprotic n of acid = (0.100)(10.00) = 1.00 mmole m = (1.00 × 10-3)(60) = 0.0600g (ii) The mass percentage of acetic acid in the solid Given: m of acid = 0.0600g, mtotal = 0.118g RTF: mass % of acid

m % = = 50.8%

d) Calculate the pH at the equivalence point of the titration of 25.0 ml 0.10M CH3COOH with 0.10 NaOH solutions. Given: = 25.0 ml, [CH3COOH] = 0.10M, [NaOH] = 0.10M, VKOH = 25.0ml, Ka = 1.80 × 10-5

28 

RTF: pH = (25.0)(0.10) = 2.5 mmoles nKOH = (25.0)(0.10) = 2.5 mmoles

CH3COOH(aq)

+ OH-(aq) → CH3COO-(aq) + H2O(l)

Initially 2.5mmoles 2.5mmole - After reaction

- - 2.5mmoles

[CH3COO-] = = 0.05M After the reaction we end up with a solution containing a weak base CH3COO-

(aq).

CH3COO-

(aq) + H2O(l) CH3COOH(aq) OH-(aq)

[initially] 0.05M - - [change] -x +x +x [eqbm 0.05 – x x x

Kb = = = 5.56 × 10-10 assume x << 0.050

[OH-] = x = = 5.27 × 10-6M (assumption valid) pOH = 5.28 pH = 14 – pOH = 14 – 5.28 = 8.72

e) The pKa values for several indicators are given below. Which of the indicators listed is most suitable for this titration? Justify your answer.

Indicator pKa Erythrosine 3 Litmus 7 O-cresolphtalein 9.2 Thymolphtalein 10

Indicators change color over a pH range = pKa ± 1 Erythrosine 2 – 4 Litmus 6 – 8 o-Cresolphthalein 8.2 – 10.2 Thymolphthalein 9 – 11

29 

Since this a titration of a weak acid with a strong base the color will change at the pH > 7, (pH at the equivalence point = 8.57). Therefore, o-Cresolphthalein is the best indicator because 8.57 lies in the range of color change of o-Cresolphthalein.

49. A pure 15.50 g sample of the weak base methylamine, CH3NH2 , is dissolved in enough distilled water to make 1000. mL of solution.

(a) Calculate the molar concentration of the CH3NH2 in the solution. Given : m = 15.50 g , v = 1000 mL ,RTF: C Number of moles= 15.50/31 = 0.5000 mole Concentration = 0.5000/ 1.000 = 0.5000M The aqueous methylamine reacts with water according to the equation below.

CH3NH2(aq) + H2O(l) ↔ CH3NH3+(aq) + OH−(aq)

(b) Write the equilibrium-constant expression for the reaction between CH3NH2(aq)

and water. Kb = [CH3NH3

+][OH-]/[CH3NH2] (c) Of CH3NH2(aq) and CH3NH3

+(aq), which is present in the solution at the higher concentration at equilibrium? Justify your answer.

CH3NH2 is present in the solution at the higher concentration at equilibrium. Methylamine is a weak base, and thus it has a small Kb value. Therefore only partial dissociation of CH3NH2 occurs in water, and [CH3NH3

+] is thus less than [CH3NH2]. (d) A different solution is made by mixing 500. mL of 0.250 M CH3NH2 with 500.

mL of 0.100 M HCl.Assume that volumes are additive. The pH of the resulting solution is found to be 10.52. (i) Calculate the concentration of OH−(aq) in the solution.

Given: pH= 10.52 RTF: concentration of OH- ions pH = −log[H+] [H+] = 10−10.52 = 3.02 × 10−11 [OH−] = Kw /[H+ ] = = 3.31 × 10−4 M

ii) Write the net-ionic equation that represents the reaction that occurs when the CH3NH2 solution is mixed with the HCl solution.

CH3NH2 + H3O+ → CH3NH3

+ + H2O

(iii) Calculate the molar concentration of the CH3NH3+(aq) that is formed in the reaction.

30 

moles of CH3NH2 = 0.500 L × 0.250 mol = 0.125 mol moles of H3O+ = 0.500 L × 0.100 mol = 0.0500 mol moles of CH3NH3+(aq) = 0.0500 mol concentration of CH3NH3+(aq) =0.0500 M

(iv) Calculate the value of Kb for CH3NH2.

[CH3NH2] = 0.0625 / 1.00 L = 0.625 M

Kb =[CH3NH3+ ][OH- ]/[CH3NH2 ] = 2.65x 10-4M

50. We need to prepare a buffered solution of pH = 5.00. Which of the following acids is best

to use? [-A-] H2C2O4 Ka = 5.9 × 10 –2 [-B-] H3AsO4 Ka = 5.6 × 10 –3 [-C-] HC2H3O2 Ka = 1.8 × 10 –5 [-D-] HOCl Ka = 3.0 × 10 –8 [-E-] HCN Ka = 4.9 × 10 –10

51. The above diagram represents the pH titration curve of a weak monoprotic acid with a

0.100 M NaOH. Use fig. 1 to answer the following questions. a. The pH at this point is less than 6. A

b. The pH at this point is greater than 8 and less than 10. C c. The pH at this point could be used to determine the acid dissociation constant. B d. What part of the curve corresponds to the optimum buffer action? B

e At this point, the solution is buffered. B f. Of the points shown on the graph, this is the point when the solution is most basic.

D g. Which of the following indicator is the best choice for the titration?

31 

Methyl orange 3.2 – 4.4 Methyl red 4.8 – 6.6 Bromothymol blue 6.1 – 7.6 Phenol phthalein 8.2 – 10.0 Alizarin 11.0 – 12.4

52. Calculate the Ksp of Al(OH)3 given that its solubility is 1.0 x 10 –11M.

Given: solubility = 1.0 x 10-11 RTF: Ksp

Al(OH)3(s) Al+3(aq) + 3OH-(aq)

[initially] --- - - [change] -s +s +3s [eqbm] -- s 3s

Ksp = [Ag+][OH-]3 = (s)(3s)3 = 27s4 = 27(1.0 x 10-11)4

Ksp = 2.7 x 10-43

53. The Ksp of Mg(OH)2 is 4.0 x 10-12. Calculate the solubility of Mg(OH)2 in water. Given: Ksp = 4.0 x 10-12 RTF: solubility

Mg(OH)2(s) Mg2+(aq) + 2OH-(aq)

[initially] --- - - [change] -s +s +2s [eqbm] -- s 2s

Ksp = [Mg2+][OH-]2 = (s)(2s)2 = 4s3 = 4.0 x 10-12 s = 1.0 x 10-4 M

54. The Ksp of AgCl is 1.0 x 10-10. Calculate the maximum mass of AgCl that will dissolve in 500cm3 solution.

Given: Ksp = 1.0 x 10-10, V = 500ml RTF: mass of AgCl that dissolves

AgCl(s) Ag+(aq) + Cl-(aq)

[initially] --- - - [change] -s +s +s [eqbm] -- s s

Ksp = [Ag+][Cl-] = (s)(s) = s2 = 1.0 x 10-10 s = 1.0 x 10-5 M

32 

⇒ 1.0 x 10-5 moles dissolve in 1000ml solution ? moles dissolve in 500mL solution ? = 5.0 x 10-6 mole

n = ⇒ m = nM = (5.0 x 10-6)(143.5) = 7.18 x 10-4 g

55. The Ksp of PbSO4 is 1.0 x 10-8. Calculate the solubility of PbSO4 in 0.01M Pb(NO3)2(aq) solution.

Given: Ksp = 1.0 x 10-8, [Pb(NO3)2] = 0.01M RTF: solubility

PbSO4(s) Pb+2(aq) + SO4-2(aq)

[initially] --- 0.01 - [change] -s +s +s [eqbm] -- 0.01 + s s

Ksp = [Pb+2][SO4

-2] = (s)(s+0.01) = 1.0 x 10-8 assume s << 0.01M s = 1.0 x 10-6 M (valid assumption)